An aircraft flies a certain distance on a bearing of 135 degree and then twice the distance on a bearing of 225 degree .it's distance from the starting point is then 350 km.find the length of the first part of the journey.WITH MORE INFORMATION PLZ!!!!!!!!!!!!!!!!!!!!!!!!!!
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Step-by-step explanation:
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Answer:
Step-by-step explanation:
t the plane start at A and go to B which is 135° position
Let AB= X
From B let it go to C in a direction BC which at position 225°
therefore ∠ABC= 225-135= 90°
BC = 2X ( Given)
So in righ angle triangle ABC
AC^2= AB^2 + BC^2
350^2 = X^2+ (2X)^2
⇒X= 70*70*5
X = 70√5 m
the length of the first part of the journey = X= 70√5 m
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